You can always let an element do two jobs at once in this sense unless you’re checking a statement that explicitly rules out the possibility, and here you are not. Here, for instance, you have to check a requirement that $d(x,z)\le d(x,y)+d(y,z)$ whenever $x,y,z\in X$; there is no implication here that $x,y$, and $z$ must be distinct points. Indeed, the triangle inequality must hold whether they are distinct of not. The same goes for the other clauses of the definition.
I already showed in this answer that $X'$ relatively compact implies $X'$ relatively countably compact. Also, if $X'$ is relatively compact, the $\overline{X'}$ is compact, and thus sequentially compact (this holds in particular in metric spaces, but also more broadly). So any sequence from $X'$ has a convergent subsequence with limit in $\overline{X'}$, so $X'$ is then relatively sequentially compact as well.
In any space, $X'$ relatively sequentially compact implies $X'$ relatively countably compact: any infinite subset $A$ of $X'$ contains some sequence with all different elements, which has a convergent subsequence to some $x \in X$, and this $x$ is an accumulation point of $A$.
If $X'$ is relatively countably compact, and $X$ is metric (first countable and $T_1$ will already do), let $(x_n)$ be a sequence from $X'$. If $A = \{x_n: n \in \mathbb{N}\}$ is finite, some value occurs an infinite number of times, and yields a convergent subsequence. So assume $A$ is infinite, so it has an accumulation point $p \in X$. Because $X$ is $T_1$, this means that every neighbourhood of $p$ intersects $A$ in infinitely many points. Pick $x_{n_1}$ in $B(p, 1)$, $x_{n_2}$ with $n_2 > n_1$ in $B(p, \frac{1}{2})$, and so on, by recursion. This defines a convergent subsequence of $(x_n)$ that converges to $p$. So $X'$ is relatively countably compact.
If $X'$ is relatively countably compact, and $X$ is metric, then we do get that $X'$ is relatively compact (this is due to Hausdorff, IIRC). I cannot reconstruct a proof right away, but there is one here, e.g. (but this involves Cauchy filters etc.)
It turns out that these notions are also equivalent in some topological vector spaces (spaces of the form $C_p(X)$ where $X$ is compact (Grothendieck), and weak topologicals on normed vector spaces (Eberlein-Smulian)), but these are a bit more involved, I think.
Best Answer
This is a bit tricky. A countable set is just that: a set. A metric space is a set equipped with some distance function. What you have shown is that for any countable set, there exists some metric on that countable set. But there are plenty of metrics on that set, and not all of them are "equivalent" (homeomorphic for example).
For example you could have put the discrete metric on your set by letting $d(x,x) = 0$ and $d(x,y) = 1$ for $x \neq y$. This satisfies all the axioms of a metric. In fact this works for any set, not only countable ones. But you could also have chosen a bijection between your set and $\{0\} \cup \{1/n \mid n \ge 1\}$, and use this bijection to define a metric. The possibilities are endless.
So to reiterate, I think the key point is: a set is just a set. Any set (including countable ones) can be equipped with some metric. Different metrics yield different metric spaces. So it's not really fair to say that "any countable set is a metric space": what is true is that any countable set can be equipped with some metric.